Hi, please advice me on implicit dereference of the shift if it is possible. 
$a = 5;
$add2= sub{ $b = shift; $$b+=2;};
&$add2(\$a);

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-this works just fine, but is there any chance to dereference it somehow as ${$shift}+=2; ? without interim variable?
!update
Mr LanX thank you very much for provided answers.
the answers are:
${shift()}+=2
${+shift}+=2
though I'm going to google what that "+" stuff exactly does.

In reply to shift implicit dereference by Lennotoecom

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